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A force F = (40 î + 10 ĵ) N acts on a body of mass 5 kg. If the body starts from rest, its position vector r at time t = 10 s, will be:

1. (100 î + 400 ĵ) m
2. (100 î + 100 ĵ) m
3. (400 î + 100 ĵ) m
4. (400 î + 400 ĵ) m

Correct answer: (400 î + 100 ĵ) m

Solution

a = F/m = (8,2) m/s^2. From rest, r = (1/2)a t^2 = 50*(8,2) = (400,100) m.

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